Question

1. Women’s Heights

• Assume that Women’s heights are normally distributed with mean μ=63.6 in. and standard deviation σ=2.5 in.

Use StatKey to answer the following questions. Include a screenshot from StatKey for each question.

1. Find the percent of women with heights between 58.6 and 68.6 inches.

2. Find the percent of women with heights between 60 inches and 65 inches.

3. Find the height of a woman in the 95^th percentile, (taller than 95% of other women.)

1. Life Expectancy Part 4

From the AllCountries data, do your best to randomly select 10 of the 213 life expectancies listed. (You can use your sample from the previous graded problem if you’d like.)

this is the 10 life expectancy samples I chose

Bermuda: 80.6

Bulgaria: 74.5

Egypt: 71.1

Korea Rep: 81.5

Argentina: 76.2

Panama: 77.6

Canada: 81.4

Korea Dem. Rep: 69.8

Belarus: 72.5

Belize: 73.9

1. Find a 90% bootstrap confidence interval for life expectancy in StatKey by taking 1000 bootstrap samples of size 10 from the population. Include a screenshot from StatKey below. What shape is the distribution? Is the distribution approximately Normal?
2. Using the estimated values for the mean and standard deviation from Part A, use a Normal distribution model to find a 90% confidence interval for life expectancy. Include a screenshot from StatKey below.
3. Explain why the results from Parts A & B are different.

Answer 1

Answer: Given that M = 63-6 & 0-2.5 a) P(58.6<x< 68:6) 020 = P(58.67 63 64 X2636-636 OS 010 OOS 586 68.6 =P[-2<Z<2 :P(Z <2)-P[22-2] = 29280.9772 -0.0228 - = 0.9544 --P ( 586<x< 68-6) - 0.9544) by PC 602X265) (60-63.6 65-63.6 -0.05 56 58 60 62 64 66 68 70 72 - P Х-д 2.5 < 2.5 PP-1.442Z<0.56) P(Z <0.56]- P(Z <= 1:44] 0.7123 -0.0749 = 0.6374 :./P (602x265] = 0.6374) 60 65 54 56 58 60 62 64 66 c) Z Score for 95% is 1.645 Azbest = OBA, AREOBUS So i xa al + Z * 63-6 +1.645*2.5 = 63.6+4.1125 = 67.7125 x = 67.7125 Thank you